Question

At what angle must the two forces (x + y) and (x – y) act so that the resultant may be $\sf{\sqrt{{x}^{2}+{y}^{2}}}$ ?

(a) $\sf{{cos}^{-1} { \bigg[}- \dfrac{( {x}^{2} + {y}^{2} ) }{2( {x}^{2} - {y}^{2} )} }{ \bigg]}[\tex]\ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ (b) [tex]\sf{ {cos}^{ - 1}{ \bigg[} - 2 \dfrac{( {x}^{2} - {y}^{2}) }{( {x}^{2} + {y}^{2}) }}{ \bigg]}$

(c) $\sf{ {cos}^{ - 1}{ \bigg[} - \dfrac{( {x}^{2} + {y}^{2}) }{( {x}^{2} - {y}^{2}) }}{ \bigg]}$

(d) $\sf{ {cos}^{ - 1}{ \bigg[} - \dfrac{( {x}^{2} - {y}^{2}) }{( {x}^{2} + {y}^{2}) }}{ \bigg]}$​

Answer 1

Answer :

$\sf Option \ (a): cos^{-1}\Bigg[-\dfrac{\left(x^2 + y^2\right)}{2\left( x^2 - y^2 \right) \ \ }\Bigg]$

Explanation :

ATQ, The two forces acting are (x + y) & (x - y). Let us name them A & B respectively. And, we know that the resultant is $\sf \sqrt{x^2 + y^2}$ .

We are asked to find the angle formed by the vectors/forces. We know that if θ is the angle formed between two vectors, then the resultant of these vectors can be found out using the below formula.

$\longmapsto \sf R^2 = A^2 + B^2 + 2ABcos\theta$

Now, substituting the forces in the equation we get,

$\Longrightarrow \sf R^2 = A^2 + B^2 + 2ABcos\theta \\ \\ \\\Longrightarrow \sf \left(\sqrt{x^2 + y^2} \ \right)^2 = \Big(x + y\Big)^2 + \Big(x - y\Big)^2 + 2\Big(x+y\Big) \Big(x-y\Big)cos\theta$

Using the below identities we get;

(a + b)² = a² + b² + 2ab

(a - b)² = a² + b² - 2ab

(a + b) (a - b) = a² - b²

$\Longrightarrow \ \sf x^2 + y^2 = x^2 + y^2 + 2xy + x^2 + y^2 - 2xy + 2cos\theta(x^2 - y^2)$

Cancelling x² + y² on both sides, and cancelling 2xy & -2xy on the right hand side we get,

$\Longrightarrow \ \sf 0 = x^2 + y^2 + 2cos\theta(x^2 - y^2)\\ \\ \\\Longrightarrow \ \sf -(x^2 + y^2) = 2cos\theta(x^2 - y^2)\\ \\ \\\Longrightarrow \ \sf 2cos\theta(x^2 - y^2) = -(x^2 + y^2)\\ \\ \\\Longrightarrow \ \sf 2cos\theta = \dfrac{-(x^2 + y^2)}{ \ \ (x^2 - y^2)}\\ \\ \\\Longrightarrow \ \sf cos\theta = \dfrac{-(x^2 + y^2)}{ \ 2(x^2 - y^2)}\\ \\ \\\Longrightarrow \ \sf \theta = cos^{-1} \ \Bigg[\dfrac{-(x^2 + y^2)}{ \ 2(x^2 - y^2)}\Bigg]$

Therefore, the answer is Option (a).

Answer 2

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$\sf{ Option \ (a): cos^{-1}\Bigg[-\dfrac{\left(x^2 + y^2\right)}{2\left( x^2 - y^2 \right) \ \ }\Bigg]}$

$\sf{ (√x² + y²)² = (x + y)² + (x − y)² + 2(x+y)(x - y) cosø }$

$\sf{ x² + y² = x² + y² + 2xy + x² + y² − }$

$\sf{ 2xy + 2(x² - y²) cosø }$

$\sf{ 2(x² - y²).cos0 = −(x² + y²) }$

$\sf{ ⇒ \: 0 \: = cos \dfrac{ -(x² + y²)}{2(x² - y²)} }$

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